# Tag Info

## New answers tagged collision

-1

Schroedinger solved this problem in 1927. The trick is to realize that in quantum mechanics, momentum is proportional to wave number (or inverse wavelength). So the COM system for an electron and a photon is the system where they have the same wavelength. The beauty of doing this in a COM system is that you have the incoming electron and the outgoing ...

0

Also, it's worth noting that the quantum mechanical model places the upper limit of the radius of an electron at 2.82e-22, which means that the electrical potential energy that two electrons would be at when they collide (given by k*q1*q2 / r) is in the order of magnitude of 50,000,000 MeV. The mass of an electron, in MeV according to e=mc^2, is closer to ...

2

The process could in general take place. A sample diagram is: $\hspace{4cm}$ With regards to Parity: Assuming the electron-positron pair don't have any angular momentum, the initial Parity is $-1$. Assuming the $\eta_C\eta_C$ pair don't have any angular momentum, their Parity is $+1$. Thus in this case the reaction cannot occur. If we assume the ...

0

If you consider things on a microscopic level, the first thing you see are the atoms at the surface of the meteor colliding with the atoms at the surface of whatever it hits, lets call it "the ground". There will be a lot of these initial collisions, each one accelerating the atoms at the surface of the ground material. These atoms in turn accelerate the ...

2

The answer is a definite maybe! Ignoring air resistance the velocity a falling object hits the ground can be calculated using the appropriate SUVAT equation or by equating potential energy lost with kinetic energy gained. However this only tells you the speed the object hits the ground, and how hard it hits the ground depends on how fast it decelerates. If ...

1

It is not random. If the exact same meteor strikes with the exact same properties then the results will be mostly the same (barring a small amount of chaos). So what is the distribution of energy. Well if you drop a ball on carpet and on a wooden floor it will make different sounds and it will bounce differently. So it is in the details of all possible ...

4

To stop instantly, you would need infinite deceleration. This in turn, requires infinite force, as demonstrable with this equation: $$\vec F=m\vec a$$ So when you hit a wall, you do not instantly stop (e.g. the trunk of the car will still move because the car is getting crushed). In a case of a change in momentum, $m\vec v$, we can use the following equation ...

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